Let \(L^2=L^2(\mathbb R^d)\) be the real Hilbert space with inner product \[ (f,g)=\int_{\mathbb R^d}f g\,dm. \]
Prove that if \(f_n\rightharpoonup f\) weakly in \(L^2\) and \[ \|f_n\|_2\to\|f\|_2, \] then \(f_n\to f\) strongly in \(L^2\).
Give a sequence of bounded functions in \(L^2\) which converges weakly along a subsequence but not strongly. What does this imply about the unit ball of \(L^2\) in the norm topology?
Solution
1. Weak convergence plus convergence of norms implies strong convergence.
Weak convergence gives, taking the test function \(g=f\), \[ (f_n,f)\longrightarrow(f,f)=\|f\|_2^2. \] Therefore \[ \begin{aligned} \|f_n-f\|_2^2 &=\|f_n\|_2^2+\|f\|_2^2-2(f_n,f)\\ &\longrightarrow \|f\|_2^2+\|f\|_2^2-2\|f\|_2^2=0. \end{aligned} \] Hence \(f_n\to f\) strongly in \(L^2\).
2. A bounded weakly convergent sequence need not converge strongly.
For \(n\ge1\), let \[ Q_n=[n,n+1]\times[0,1]^{d-1} \] and define \[ e_n=\chi_{Q_n}. \] Then each \(e_n\) is bounded by \(1\) and \[ \|e_n\|_2=1. \] The sets \(Q_n\) are pairwise disjoint, so \((e_n)\) is an orthonormal sequence. Hence for every \(g\in L^2\), \[ (e_n,g)\to0 \] by Bessel’s inequality. Thus \[ e_n\rightharpoonup0. \] But \[ \|e_n-0\|_2=1 \] for every \(n\), so the sequence does not converge strongly. In fact, for \(m\ne n\), \[ \|e_n-e_m\|_2=\sqrt2, \] so no subsequence is strongly Cauchy.
Therefore the closed unit ball of \(L^2(\mathbb R^d)\) is not compact in the norm topology.